10/9/2013

Today:

Schedule an appointment or stop
by office hours to pick up QUIZ 2

Next Meeting
◦ Stoichiometric Analysis:
 Mole to Mole Conversions:
 Use stoichiometric coefficients
from balanced equations
◦ Concept Check:
 Covers today’s material
 Gram to Gram Conversions:
◦ Check out Canvas for
Stoichiometry Practice Problems
 Use MOLAR MASS to get to moles
 Theoretical Yield:
◦ Reading for Wednesday:
 How much product can form in a
reaction?
 Start Chapter 9, pp. 285-305
 Limiting Reagents:
 Method 1
 Method 2
Graded Concept Check:
Decomposition of Sodium Azide
A.
Sodium azide (NaN3) decomposes to produce sodium metal &
nitrogen gas. What is the balanced chemical equation for this
reaction?
NaN3(s)
Na(s) + 3 N(g)
B.
NaN3(s)
Na(s) + N3(g)
C.
2 NaN3(s)
2 Na(s) + 3 N2(g)
D.
2 NaN3(s)
Na2(s) + 3 N2(g)
E. 2 NaN3(s) + 2 O2(g)
Na2O(s) + 3 N2O(g)
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10/9/2013
Stoichiometry:
Studying the quantities in chemical reactions
Law of Conservation of Mass: requires the same number
of atoms on each side of the chemical equation.

Helpful in predicting the amount of
products that can form based on the
amount of starting reactants
2 NaN3(s)

2 Na(s) + 3 N2(g)
2 sodium azide units will decompose
into just 2 sodium atoms & 3
nitrogen molecules.
Sodium azide decomposition is
used to inflate air bags.
Mole-to-Mole Conversions:
Stoichiometric coefficients express PARTICLE ratios
2 NaN3(s)
2 Na(s) + 3 N2(g)
How many moles of N2 could form from 6 moles of NaN3?
Theoretical Yield: MAXIMUM amount of product that can form
in a reaction based on the limited starting materials.
How many moles of NaN3 are needed to produce 2.67 moles of N2 gas
(about the amount of gas needed to fill one airbag)?
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10/9/2013
iClicker Participation Question:
Mole-to-Mole Conversions—From Reactants to Products
How much NH3 could be produced, if 40 moles of hydrogen
gas reacted according to the equation below?
A. 80 moles
N2(g) + 3 H2(g)
2 NH3(g)
B. 60 moles
C. 40 moles
“The expansion of the world's population
from 1.6 billion people in 1900 to today's
seven billion [in 2012] would not have been
possible without the [industrial] synthesis
of ammonia.“ –MIT Press
D. 27 moles
E. 20 moles
Mass-to-Mass Conversions:
How many grams of sodium metal can be produced from the
decomposition of 130 grams of sodium azide?
# of grams
NaN3
2 NaN3(s)
2 Na(s) + 3 N2(g)
# of grams
Na
Moles cannot be measured directly, but…
Molar Mass
of NaN3:
22.99 +
3 x 14.01 =
65.02 g/mol
MASS can be measured on a balance &
tied to the number of moles with
MOLAR MASS
Mole-to-mole ratios relate
reactants & products:
# of moles
NaN3
2 NaN3 produce 2 Na
Molar Mass
Na:
22.99 g/mol
# of moles
Na
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10/9/2013
What volume of water could form from the
combustion of 20. mL of C3H8O?
2 C3H8O + 9 O2(g)
6 CO2(g) + 8 H2O(g)
2 mol of C3H8O
make 8 mol H2O
# of moles
C3 H 8 O
MOLAR MASS = 60.1 g/mol
# of moles
H2 O
MOLAR MASS = 18.0 g/mol
Mass of
C3 H 8 O
Mass of
H2 O
DENSITY = 0.786 g/mL
DENSITY = 1.00 g/mL
Volume of
C3 H 8 O
Volume of
H2 O
Limiting Reagents:
2 C3H8O + 9 O2(g)
Trial 1:
• For the reaction to occur, C3H8O
must combine with O2 in a 2 to
9 ratio
• 20 mL C3H8O reacts O2 in bottle
to produce CO2, H2O, & HEAT
6 CO2(g) + 8 H2O(g)
Trial 2:
•
•
•
Without
NO REACTION
2, there isReagent:
O2 is theOLimiting
It
Inside
the
bottle:
O
is
depleted
LIMITS the amount2 of product
Reaction is LIMITED to the
that can form. The other
opening where O2 is present
reactant is in EXCESS.
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10/9/2013
How many grams of oxygen would be needed to
completely react with 20. mL of C3H8O?
2 C3H8O
# of moles
C3 H 8 O
+
9 O2(g)
2 mol of C3H8O
react with 9 mol O2
MOLAR MASS = 60.1 g/mol
# of moles
O2
MOLAR MASS = 32.0 g/mol
Mass of
C3 H 8 O
Mass of
O2
DENSITY = 0.786 g/mL With LESS than 38 g O : the 20 mL of
2
C3H8O cannot fully react. This would
Volume of
make the O2 the limiting reagent &
C3 H 8 O
the C3H8O would be in excess.
Stoichiometric Coefficients:
A recipe for making (chemical) products
• The stoichiometric coefficients specify the exact PARTICLE ratio
necessary to convert the reactants to the products.
4H
4
+
+ 1
1C
1 CH4
1
It will always require 4 wheels and 1 frame to make the core structure of a car:
• 3 wheels & 1 frame could not make a complete product. In this scenario,
the wheels would be LIMITING the production of a complete car.
• 120 wheels & 20 frames: more frames would be necessary to use up all
the available wheels. The frames would be LIMITING. Without more
frames, some unused wheels would be left over.
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10/9/2013
iClicker Participation Question:
Predicting the Volume of Gases Produced
Which mixture will yield the MOST CO2 GAS?
CH3CO2H(aq) + NaHCO3(s)
A
3.0 g (0.05 mol)
4.2 g (0.05 mol)
B
6.0 g (0.10 mol)
4.2 g (0.05 mol)
C
3.0 g (0.05 mol)
8.4 g (0.10 mol)
NaCH3CO2(aq) + H2O(l) + CO2(g)
D. Each mixture will yield the same volume of gases
E. Both B & C will yield the same amount of gas,
which will be more than A
When the ions switch places, be sure
write the formulas of the products so
iClicker Participation
Question:
that NEUTRAL
compounds form (with the
Predicting the Volume
of Gases
Produced
smallest
whole
number ratio of atoms).
Which mixture will yield the MOST CO2 GAS?
CH3CO2H(aq) + NaHCO3(s)
NaCH3CO2(aq) + H2O(l) + CO2(g)
A
3.0 g (0.05 mol)
4.2 g (0.05 mol)
BOTH FULLY REACT
B
6.0 g (0.10 mol)
4.2 g (0.05 mol)
NaHCO3 is LIMITING
C
3.0 g (0.05 mol)
8.4 g (0.10 mol)
CH3CO2H is LIMITING
D. Each mixture will yield the same volume of gases
The Limiting Reagent is CONSUMED & used up first, leaving
a portion of the other reactant in EXCESS (unreacted).
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10/9/2013
Nitrogen Dioxide: A major component of smog
•
Nitrogen monoxide (aka nitric oxide, NO) is produced in internal
combustion engines.
•
Nitric oxide released into the atmosphere can react with O2 to
form nitrogen dioxide (aka nitrous oxide, NO2).
2 NO + O2
2 NO2
Determining the Limiting Reagent: METHOD 1
While calculating the THEORETICAL YIELD
1.
2.
Convert the starting quantity of each reactant to the quantity of product that
could potentially form.
The limiting reagent is the reactant that would produce LESS product and
determines the THEORETICAL YIELD.
4 mol O2
4 mol NO
O2 is in EXCESS
Nitric oxide
is LIMITING
(some remains
unreacted in the end)
(it is consumed &
limits the formation
of product)
Animations adapted from McGraw Hill Publishers
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10/9/2013
Determining the Limiting Reagent: METHOD 2
Comparing one reactant to another
1. Start with Reactant A. Given the starting quantity of A, calculate the quantity
of the other reactant (B) needed to FULLY react with A.
2. Compare the available quantity of B to the amount needed to react with A. If
more is available than is needed, B is in EXCESS & A is the LIMITING.
2 mol O2
6 mol NO
Nitric oxide
is in EXCESS
Oxygen is
LIMITING (it is
(some remains
unreacted in the end)
consumed & limits the
formation of product)
Animations adapted from McGraw Hill Publishers
A Stoichiometric Mixture:
Neither reactant is limiting or in excess
•
When the reactants are in the proper stoichiometric ratio, each can FULLY
react with the other. Neither starting material remains in the end of the
reaction. Neither is considered a limiting reagent.
3 mol O2
6 mol NO
Animations adapted from McGraw Hill Publishers
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10/9/2013
Practice Problems:
Urea (CH4N2O), a common fertilizer, can be synthesized by the reaction of
ammonia (NH3) with carbon dioxide.
2 NH3(aq) + CO2(aq)
CH4N2O(aq) + H2O(l)
An industrial synthesis of urea begins with 35.8 kg of aammonia and 89 kg of
carbon dioxide. What mass of urea could theoretically be produced from this
starting mixture?
An emergency reathing apparatus placed in mines or caves works via the
chemical reaction below:
4 KO2(s) + 2 CO2(g)
2 K2CO3 + 3 O2(g)
If the oxygen supply becomes limited, a worker can use the apparatus to
breathe while exiting the mine. Notice that the reaction produces O 2, which can
be breathed, and absorbs CO2, a product of respiration. What minimum amount
of KO2 is required for the apparatus to provide enough oxygen to allow the user
15 minutes to exit the mine? Assume 4.4 g of O2 are needed for 15 minutes of
normal breathing.
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